How do you find the square root of an imaginary number of the form a+bi?

Suppose #a+bi = (c+di)^2#

How do we solve for #c# and #d#?

#(c+di)^2 = c^2+2cdi + d^2i^2 = (c^2-d^2) + (2cd)i#

So we want to solve:

#c^2-d^2 = a#

#2cd = b#

From the second of these, we find:

#d = b/(2c)#

So:

#d^2 = b^2/(4c^2)#

Substituting this in the first equation we get:

#c^2-b^2/(4c^2) = a#

Multiply through by #4c^2# to get:

#4(c^2)^2-b^2 = 4ac^2#

Subtract #4ac^2# from both sides to get:

#4(c^2)^2-4a(c^2)-b^2 = 0#

From the quadratic formula, we find:

#c^2 = (4a+-sqrt((4a)^2+16b^2))/8 =(a+-sqrt(a^2+b^2))/2#

For #c# to be Real valued we require #c^2 >= 0#, hence we need to choose the root with the #+# sign...

#c^2 = (a+sqrt(a^2+b^2))/2#

Hence:

#c = +-sqrt((sqrt(a^2+b^2)+a)/2)#

Then:

#d = +-sqrt(c^2-a)#

#= +-sqrt((sqrt(a^2+b^2)+a)/2-a)#

#= +-sqrt((sqrt(a^2+b^2)-a)/2)#

The remaining question is: What signs do we need to choose?

Since #2cd = b#:

If #b > 0# then #c# and #d# must have the same signs.

If #b < 0# then #c# and #d# must have opposite signs.

If #b=0# then #d=0#, so we don't have to worry.

If #b != 0# then we can use #b/abs(b)# as a multiplier to match the signs as we require to find that the square roots of #a+bi# are:

#+-((sqrt((sqrt(a^2+b^2)+a)/2)) + (b/abs(b) sqrt((sqrt(a^2+b^2)-a)/2))i)#

Footnote

The question asked what is the square root of #a+bi#.

For positive Real numbers #x#, the principal square root of #x# is the positive one, which is the one we mean when we write #sqrt(x)#. This is also the square root that people commonly mean when they say "the square root of #2#" and suchlike, neglecting the fact that #2# has two square roots.

For negative Real numbers #x#, then by convention and definition, #sqrt(x) = i sqrt(-x)#, That is, the principal square root is the one with a positive coefficient of #i#.

It becomes more complicated when we deal with the square roots of Complex numbers in general.

Consider:

#sqrt(-1/2-sqrt(3)/2i) = +-(1/2-sqrt(3)/2 i)#

Which sign do you prefer?